2018 AMC 10A Problem 25

Every number here is a repeated digit, so write each one using a string of ones. Let $R$ be the $n$-digit number $11\ldots1$.

1. Write $A_n$ and $B_n$. $A_n = aR$ and $B_n = bR$. For example, with $n = 2$, $66 = 6 \cdot 11$.
2. Write $C_n$. A $2n$-digit string of ones is $R$ followed by $R$, which is $R \cdot (10^n + 1)$. For example, $1111 = 11 \cdot 101$. Also $10^n + 1 = 9R + 2$ (check: $101 = 9 \cdot 11 + 2$). So
$$C_n = c \cdot R \cdot (9R + 2).$$
3. Put everything into $C_n - B_n = A_n^2$.
$$cR(9R + 2) - bR = a^2 R^2.$$
Every term has a factor $R$, so divide it out:
$$9cR + 2c - b = a^2 R.$$
Move the $R$ terms to one side:
$$(9c - a^2) \cdot R = b - 2c.$$
4. Why two values of $n$ matter. Different $n$ give different $R$ ($1$, $11$, $111$, ...), but the right side $b - 2c$ stays the same. If $9c - a^2$ were not $0$, only one value of $R$ could make the two sides equal, so only one $n$ would work. For two values of $n$ we need
$$9c = a^2 \qquad\text{and}\qquad b = 2c,$$
and then the equation holds for every $n$.
5. Try the digits. $a^2 = 9c$ means $a^2$ is a multiple of $9$, so $a$ is $3$, $6$ or $9$.
- $a = 3$: $c = 1$, $b = 2$. Sum $6$.
- $a = 6$: $c = 4$, $b = 8$. Sum $18$.
- $a = 9$: $c = 9$, $b = 18$, which is not a digit.

The largest sum is $6 + 8 + 4 = 18$.
6. Check. $n = 1$: $44 - 8 = 36 = 6^2$. $n = 2$: $4444 - 88 = 4356 = 66^2$.

The answer is $\boxed{\textbf{(D)}\ 18}$.

Numbers made of one repeated digit are that digit times a string of ones, and a string of $2n$ ones is built from the string of $n$ ones. That turns the whole problem into one equation in $R$. The equation has the form (number) times $R$ = (number), and if it must hold for two different $R$, both numbers must be $0$. "At least two values of $n$" is the hint to use this.

Write out $n = 1$ and $n = 2$ to see where the conditions come from. For $n = 1$: $\overline{cc} - b = a^2$, so $a^2 + b = 11c$. For $n = 2$: $1111c - 11b = 121a^2$, so dividing by $11$, $11a^2 + b = 101c$. Multiply the first equation by $11$ and subtract the second: $10b = 20c$, so $b = 2c$, and then $a^2 = 9c$. This shows what happens when $n = 1$ and $n = 2$ both work; the main solution shows the same must happen for any two values of $n$.